SATHEE: Work Power and Energy 3 Question 7

Work Power and Energy 3 Question 7

7. Two blocks $A$ and $B$ are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown in the figure. Block $B$ slides over the horizontal top surface of a stationary block $C$ and the block $A$ slides along the vertical side of $C$ , both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2 . Force constant of the spring is $1960 N / m$ . If mass of block $A$ is $2 k g$ . Calculate the mass of block $B$ and the energy stored in the spring.

(1982, 5M)

Show Answer

Answer:

Correct Answer: 7. $10 k g, 0.098 J$

Solution:

Normal reaction between blocks $A$ and $C$ will be zero. Therefore, there will be no friction between them.

Both $A$ and $B$ are moving with uniform speed. Therefore, net force on them should be zero.

For equilibrium of $A$

$\begin{aligned} m_{A} g & = k x \\ ∴ x & = \frac{m_{A} g}{k} = \frac{(2) (9.8)}{1960} \\ = 0.01 m \end{aligned}$

For equilibrium of $B$

$\begin{aligned} μ m_{B} g = T & = k x = m_{A} g \\ ∴ m_{B} & = \frac{m_{A}}{μ} = \frac{2}{0.2} \\ = 10 k g \end{aligned}$

Energy stored in spring

$\begin{aligned} U & = \frac{1}{2} k x^{2} = \frac{1}{2} (1960) (0.01)^{2} \\ = 0.098 J \end{aligned}$

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